Top person sorted by score

At this site we keep several lists of primes, most notably the list of the 5,000 largest known primes. Who found the most of these record primes? We keep separate counts for persons, projects and programs. To see these lists click on 'number' to the right.

Clearly one 100,000,000 digit prime is much harder to discover than quite a few 100,000 digit primes. Based on the usual estimates we score the top persons, provers and projects by adding ‎(log n)³ log log n‎ for each of their primes n. Click on 'score' to see these lists.

Finally, to make sense of the score values, we normalize them by dividing by the current score of the 5000th prime. See these by clicking on 'normalized score' in the table on the right.

rank person primes score 21 Arno Lehmann 3 52.2822 22 Sylvanus A. Zimmerman 4 52.2575 23 Stefan Larsson 199 52.2127 24 Ben Maloney 1 52.0371 25 Frank Matillek 10 52.0287 26 Wolfgang Schwieger 94 51.9680 27 Marc Wiseler 9 51.8176 28 Diego Bertolotti 1 51.6397 29 Rudi Tapper 4 51.6208 30 Brian D. Niegocki 26 51.4097 31 Randall Scalise 150 51.3982 32 Hiroyuki Okazaki 55 51.2916 33 Antonio Lucendo 20 51.1797 34 Valter Cavecchia 45 51.1745 35 Vaughan Davies 55 51.0205 36 Peter Kaiser 81.3333 51.0002 37 Erik Veit 52 50.9911 38 Alen Kecic 18 50.9744 39 Detlef Lexut 15 50.9555 40 Michael Cameron 1 50.9234

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Notes:

Score for Primes

To find the score for a person, program or project's primes, we give each prime n the score (log n)³ log log n; and then find the sum of the scores of their primes. For persons (and for projects), if three go together to find the prime, each gets one-third of the score. Finally we take the log of the resulting sum to narrow the range of the resulting scores. (Throughout this page log is the natural logarithm.)

How did we settle on (log n)³ log log n? For most of the primes on the list the primality testing algorithms take roughly O(log(n)) steps where the steps each take a set number of multiplications. FFT multiplications take about

O( log n ^. log log n ^. log log log n )

operations. However, for practical purposes the O(log log log n) is a constant for this range number (it is the precision of numbers used during the FFT, 64 bits suffices for numbers under about 2,000,000 digits).

Next, by the prime number theorem, the number of integers we must test before finding a prime the size of n is O(log n) (only the constant is effected by prescreening using trial division).  So to get a rough estimate of the amount of time to find a prime the size of n, we just multiply these together and we get

O( (log n)³ log log n ).

Finally, for convenience when we add these scores, we take the log of the result.  This is because log n is roughly 2.3 times the number of digits in the prime n, so (log n)³ is quite large for many of the primes on the list. (The number of decimal digits in n is floor((log n)/(log 10)+1)).